Theorem syl6rbb 415

Description: A syllogism inference from two biconditionals.

Hypotheses

Ref	Expression
syl6rbb.1	⊢ (φ → (ψ ↔ χ))
syl6rbb.2	⊢ (χ ↔ θ)

Assertion

Ref	Expression
syl6rbb	⊢ (φ → (θ ↔ ψ))

Proof of Theorem syl6rbb

Step	Hyp	Ref	Expression
1		syl6rbb.1	. . 3 ⊢ (φ → (ψ ↔ χ))
2		syl6rbb.2	. . 3 ⊢ (χ ↔ θ)
3	1, 2	syl6bb 414	. 2 ⊢ (φ → (ψ ↔ θ))
4	3	bicomd 399	1 ⊢ (φ → (θ ↔ ψ))

Syntax hints:   → wi 2   ↔ wb 127

This theorem is referenced by:  syl6rbbr 417  muln0bt 4213  elznn0 4576  norm-it 5080

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6

This theorem depends on definitions:  df-bi 128  df-an 198

metamath.org